Problem: Solve for $x$ : $ 7|x - 1| - 7 = -4|x - 1| + 7 $
Answer: Add $ {4|x - 1|} $ to both sides: $ \begin{eqnarray} 7|x - 1| - 7 &=& -4|x - 1| + 7 \\ \\ { + 4|x - 1|} && { + 4|x - 1|} \\ \\ 11|x - 1| - 7 &=& 7 \end{eqnarray} $ Add ${7}$ to both sides: $ \begin{eqnarray} 11|x - 1| - 7 &=& 7 \\ \\ { + 7} &=& { + 7} \\ \\ 11|x - 1| &=& 14 \end{eqnarray} $ Divide both sides by ${11}$ $ \dfrac{11|x - 1|} {{11}} = \dfrac{14} {{11}} $ Simplify: $ |x - 1| = \dfrac{14}{11}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 1 = -\dfrac{14}{11} $ or $ x - 1 = \dfrac{14}{11} $ Solve for the solution where $x - 1$ is negative: $ x - 1 = -\dfrac{14}{11} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& -\dfrac{14}{11} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& -\dfrac{14}{11} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $11$ $ x = - \dfrac{14}{11} {+ \dfrac{11}{11}} $ $ x = -\dfrac{3}{11} $ Then calculate the solution where $x - 1$ is positive: $ x - 1 = \dfrac{14}{11} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& \dfrac{14}{11} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& \dfrac{14}{11} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $11$ $ x = \dfrac{14}{11} {+ \dfrac{11}{11}} $ $ x = \dfrac{25}{11} $ Thus, the correct answer is $x = -\dfrac{3}{11} $ or $x = \dfrac{25}{11} $.